2 years ago

Please help!

Mathematics

1 answer:

wariber [46]2 years ago

5 0

Ya is = 10x-28 see the picture for explanation

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HEY YOU! STOP YOUR SCROLLING AND HELP ME OUT?! PLEASE?!

rusak2 [61]

t=2 should be your answer!

7 0

3 years ago

Read 2 more answers

Find the value or measure. Assume all lines that appear to be tangent are tangent. X=

aev [14]

According to the secant-tangent theorem, we have the following expression:

$(x+3)^2=10.8(19.2+10.8)$

Now, we solve for <em>x</em>.

$\begin{gathered} x^2+6x+9=10.8(30) \\ x^2+6x+9=324 \\ x^2+6x+9-324=0 \\ x^2+6x-315=0 \end{gathered}$

Then, we use the quadratic formula:

$x_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}$

Where a = 1, b = 6, and c = -315.

$\begin{gathered} x_{1,2}=\frac{-6\pm\sqrt[]{6^2-4\cdot1\cdot(-315)}}{2\cdot1} \\ x_{1,2}=\frac{-6\pm\sqrt[]{36+1260}}{2}=\frac{-6\pm\sqrt[]{1296}}{2} \\ x_{1,2}=\frac{-6\pm36}{2} \\ x_1=\frac{-6+36}{2}=\frac{30}{2}=15 \\ x_2=\frac{-6-36}{2}=\frac{-42}{2}=-21 \end{gathered}$ <h2>Hence, the answer is 15 because lengths can't be negative.</h2>